Question

A block of mass M has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially the right edge of the block is at x = 0, in a co-ordinate system fixed to the table. A point mass m is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block, its position is x and the velocity is v. At that instant, which of the following options is/are correct?

(A) The x component of displacement of the center of mass of the block M is : \(-\frac{mR}{M+m}\)

(B) The position of the point mass is : x = \(-\sqrt{2}\frac{mR}{M+m}\)

(C) The velocity of the point mass m is : v = \(\sqrt{\frac{2gR}{1+\frac{m}{M}}}\)

(D) The velocity of the block M is: V = \(-\frac{m}{M}\sqrt{2gR}\)

Answer 1

 (A) The x component of displacement of the center of mass of the block M is : \(-\frac{mR}{M+m}\)

 (C) The velocity of the point mass m is : v = \(\sqrt{\frac{2gR}{1+\frac{m}{M}}}\)

Δ x_cm of the block & point mass system = 0 

∴ m (x + R) + Mx = 0

where x is displacement of the block.

x = -\(\frac{mR}{M+m}\)

From conservation of momentum and mechanical energy of the combined system 

0 = mv – MV